`int_(0)^(2) |(1-x)|dx`=

Options

A. 1
B. 2
C. 4
D. none

Answer

To solve the integral \( \int_{0}^{2} |1 - x| \, dx \), we will break it down into two parts based on the behavior of the absolute value function.

### Step 1: Identify where the expression inside the absolute value changes sign
The expression \( |1 - x| \) changes sign at \( x = 1 \). Therefore, we will split the integral into two parts:
- From \( 0 \) to \( 1 \), \( 1 - x \geq 0 \) so \( |1 - x| = 1 - x \).
- From \( 1 \) to \( 2 \), \( 1 - x < 0 \) so \( |1 - x| = -(1 - x) = x - 1 \).

### Step 2: Set up the integral
We can now express the integral as:
\[ \int_{0}^{2} |1 - x| \, dx = \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx \]

### Step 3: Calculate the first integral
Calculate \( \int_{0}^{1} (1 - x) \, dx \):
\[ \int (1 - x) \, dx = x - \frac{x^2}{2} + C \]
Now, evaluate from \( 0 \) to \( 1 \):
\[ \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1^2}{2} \right) - \left( 0 - 0 \right) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 4: Calculate the second integral
Now calculate \( \int_{1}^{2} (x - 1) \, dx \):
\[ \int (x - 1) \, dx = \frac{x^2}{2} - x + C \]
Evaluate from \( 1 \) to \( 2 \):
\[ \left[ \frac{x^2}{2} - x \right]_{1}^{2} = \left( \frac{2^2}{2} - 2 \right) - \left( \frac{1^2}{2} - 1 \right) = \left( 2 - 2 \right) - \left( \frac{1}{2} - 1 \right) = 0 - \left( \frac{1}{2} - 1 \right) = 0 + \frac{1}{2} = \frac{1}{2} \]

### Step 5: Combine the results
Now combine the results of the two integrals:
\[ \int_{0}^{2} |1 - x| \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{2} |1 - x| \, dx \) is \( \boxed{1} \).